what is the conjugate of 1/(1-i)
correction \[1/(i-1)\]
First simplify it. Get rid of the fractional form Then finding the conjugate is easy
I need the conjugate to get rid of the i in the denominator
No you only need the conjugate of the denominator for that.
I'm sorry that is what I am confused on
You have this fraction whose conjugate you want. First you want to make the fraction have the form a+bi so that the conjugate is a-bi. To make the fraction have the form a+bi you need to get rid of the i in the denominator. So you multiply the numerator and denominator by the conjugate of the denominator to make the denominator real
First multiply both numerator and denominator by i+1 . it will give you (i+1)/(i2-1) . Since i2=-1 so it simplifies to (i+1)/-2 whose conjugate will be (1-i)/-2
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